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## Regular and effective monomorphisms and epimorphisms

Previously we observed that although monomorphisms tended to give expected generalizations of injective function in many categories, epimorphisms sometimes weren’t the expected generalization of surjective functions. We also discussed split epimorphisms, but where the definition of an epimorphism is too permissive to agree with the surjective morphisms in familiar concrete categories, the definition of a split epimorphism is too restrictive.

In this post we will discuss two other intermediate notions of epimorphism. (These all give dual notions of monomorphisms, but their epimorphic variants are more interesting as a possible solution to the above problem.) There are yet others, but these two appear to be the most relevant in the context of abelian categories.

Regular

Recall that the equalizer of a parallel pair of morphisms $f, g : a \to b$ is their limit. More explicitly, it is the universal object $\text{Eq}(f, g)$ together with a map $\text{eq}(f, g) : \text{Eq}(f, g) \to a$ such that $f \circ \text{eq}(f, g) = g \circ \text{eq}(f, g)$. A map with this property is said to equalize $f$ and $g$, so the equalizer is the universal equalizing map.

Dually, the coequalizer of a parallel pair of morphisms $f, g : a \to b$ is their colimit. More explicitly, it is the universal object $\text{Coeq}(f, g)$ together with a map $\text{coeq}(f, g) : b \to \text{Coeq}(f, g)$ such that $\text{coeq}(f, g) \circ f = \text{coeq}(f, g) \circ g$.

The definition of equalizer and coequalizer can in principle be extended to any family of morphisms, but generally speaking the terms “equalizer” and “coequalizer” refer to a pair of morphisms.

Example. In $\text{Set}$, the equalizer of a parallel pair of functions $f, g : X \to Y$ is $\{ x \in X : f(x) = g(x) \} \subset X$, and the coequalizer of $f, g$ is the quotient of $Y$ by the identifications $f(x) = g(x)$. Roughly speaking, equalizers cut out subsets of the source where some condition holds while coequalizers force certain conditions to hold in the target.

Sub-example. If $f : X \to X$ is an endomorphism, the equalizer of $f, \text{id}_X : X \to X$ is the set $\text{Fix}(f)$ of fixed points of $f$.

Example. Let $C$ be any concrete category whose forgetful functor $U : C \to \text{Set}$ has a left adjoint. Then $U$ preserves limits, so if equalizers exist in $C$, then on underlying sets they look like they do above. Coequalizers may look different; for example, in $\text{Grp}$, the coequalizer of a parallel pair of morphisms $f_1, f_2: G \to H$ is the quotient of $H$ by the normal subgroup generated by the elements $f_1(g) f_2(g)^{-1}, g \in G$.

Example. The forgetful functor $\text{Top} \to \text{Set}$ has both a left and right adjoint, so preserves both equalizers and coequalizers. Consequently the above description carries over to $\text{Top}$ as well. The coequalizer in particular should be thought of as a mild generalization of the notion of a quotient space.

Example. Let $G$ be a group acting on an object $c$ in a category $C$. Then the equalizer of all of the morphisms $g : c \to c, g \in G$ describes the $G$-invariants $c^G$ of $c$, while their coequalizer describes the $G$-coinvariants $c_G$ of $c$.

Sub-example. If $C = \text{Ab}$ and $A \in \text{Ab}$ is an abelian group, then $A^G = \{ a \in A : \forall g \in G : ga = a \}$ and $A^G = A/\text{span}(ga - a : g \in G)$.

Example. In any $\text{Ab}$-enriched category, the equalizer of a parallel pair of morphisms $f, g : a \to b$ is the equalizer of $f - g, 0 : a \to b$, otherwise known as the kernel of $f - g$. In this case we write $\text{Ker}(f - g) = \text{Eq}(f, g)$ and $\text{ker}(f - g) = \text{eq}(f, g)$.

Dually, the coequalizer of a parallel pair of morphisms $f, g : a \to b$ is the coequalizer of $f - g, 0 : a \to b$, otherwise known as the cokernel of $f - g$. In this case we write $\text{Coker}(f - g) = \text{Coeq}(f, g)$ and $\text{coker}(f - g) = \text{coeq}(f, g)$.

The kernel and cokernel as defined here reproduce the usual notion of kernel and cokernel in, say, $R\text{-Mod}$ for $R$ a ring. The kernel and cokernel make sense in any category with zero morphisms, but computing kernels and cokernels is not in general equivalent to computing equalizers and coequalizers in such a category since we may not have the ability to subtract morphisms.

Definition-Proposition: The equalizer $\text{eq}(f, g) : \text{Eq}(f, g) \to a$ of some pair of morphisms is a regular monomorphism. In particular, a regular monomorphism is a monomorphism.

Proof. Let $p, q : b \to \text{Eq}(f, g)$ be a pair of morphisms such that $\text{eq}(f, g) \circ p = \text{eq}(f, g) \circ q$. Call this composite $h : b \to a$. Then $h$ equalizes $f, g$, hence factors uniquely through $\text{Eq}(f, g)$. Since $p, q$ both constitute factorizations of $h$ through $\text{Eq}(f, g)$, it follows that $p = q$.

Dually, we obtain the following.

Definition-Proposition: The coequalizer $\text{coeq}(f, g) : b \to \text{Coeq}(f, g)$ of some pair of morphisms is a regular epimorphism. In particular, a regular epimorphism is an epimorphism.

Note that, like being a monomorphism (resp. epimorphism), being a regular monomorphism (resp. regular epimorphism) is a limit property (resp. colimit property).

Regular epimorphisms should be thought of as being generalized quotient maps in a sense which will be made precise in the next section. We defer examples until the next section, where we will prove a result which will allow us to verify examples in many classes of categories. For now, consider the following.

Proposition: Let $f, g : a \to b$ be a parallel pair of morphisms and let $h : b \to c$ be a monomorphism. Then the induced map $\text{Eq}(f, g) \to \text{Eq}(h \circ f, h \circ g)$ is an isomorphism.

Proof. It suffices to verify that the two equalizers have the same universal property: a map $e : z \to a$ satisfies $f \circ e = g \circ e$ if and only if it satisfies $h \circ f \circ e = h \circ g \circ e$ since $h$ is left-cancellative. $\Box$

Dually, we obtain the following.

Proposition: Let $f, g : a \to b$ be a parallel pair of morphisms and let $h : c \to a$ be an epimorphism. Then the induced map $\text{Coeq}(f \circ h, g \circ h) \to \text{Coeq}(f, g)$ is an isomorphism.

Roughly speaking, the above propositions say that equalizers don’t change if we embed the target into something bigger and coequalizers don’t change if we think of the source as the quotient of something bigger.

Effective

The definition of a regular monomorphism (resp. epimorphism) only refers to being the coequalizer of some pair of maps. However, in order to determine whether a morphism is a regular monomorphism (resp. epimorphism), there is a natural pair of maps to consider if it exists.

Let $f : a \to b$ be a morphism in a category. The kernel pair of $f$ is the pullback of the diagram $a \xrightarrow{f} b \xleftarrow{f} a$; equivalently, it is the fiber product $a \times_b a$. Dually, the cokernel pair of $f$ is the pushout of the diagram $b \xleftarrow{f} a \xrightarrow{f} b$; equivalently, it is the amalgamated coproduct $b \sqcup_a b$.

The following examples will only discuss kernel pairs. We already saw some examples of cokernel pairs earlier when trying to determine whether various maps were epimorphisms. Kernel pairs should be thought of as internal equivalence relations or congruences associated to a map; this can be given a formal definition but we will not need it.

Example. Let $f : X \to Y$ be a map of sets. Then

$\displaystyle X \times_Y X = \{ (x_1, x_2) \in X \times X : f(x_1) = f(x_2) \}$.

In other words, as a subset of $X \times X$, it is precisely the equivalence relation induced by $f$ on $X$.

Note that the above subset or subspace does not change upon replacing the codomain by $\text{im}(f)$; in other words, to compute kernel pairs we may assume without loss of generality that $f$ is surjective. This generalizes.

Note also that if $f$ is surjective then the coequalizer of the two projections $p_1, p_2 : X \times_Y X \to X$ is $Y$. This also generalizes.

Example. Let $C$ be any concrete category whose forgetful functor $U : C \to \text{Set}$ has a left adjoint. Then $U$ preserves limits, so if kernel pairs exist in $C$, then on underlying sets they look like they do above. This is true in particular for $C = \text{Grp}$. Here an alternate interpretation of the kernel pair of a group homomorphism $f : G \to H$ with kernel $N$ is that it consists of the subgroup of pairs $(g_1, g_2) \in G \times G$ such that $g_1 \equiv g_2 \bmod N$; similarly, if $f : R \to S$ is a ring homomorphism with kernel $I$, then the kernel pair consists of the subgroup of pairs $(r_1, r_2) \in R \times R$ such that $r_1 \equiv r_2 \bmod I$.

Recall that in $\text{Set}$ there is a natural way to pass back and forth between equivalence relations on a set $X$ and surjective functions $f : X \to Y$ with source $X$. One direction of this equivalence is given by taking the kernel pair of $f$. This produces a subset $R = X \times_Y X$ of $X \times X$ together with a pair of projections $p_1, p_2 : R \to X$. The other direction of this equivalence is given by taking the coequalizer of $p_1, p_2$. This suggests the following definition.

Definition: A morphism $f : a \to b$ is an effective epimorphism if it has a kernel pair and is the coequalizer of its kernel pair.

(Note that the data of the kernel pair includes the data of a pair of maps to $a$, and as above it is this pair of maps that we take the coequalizer of.)

Dually, we obtain the following definition.

Definition: A morphism $f : a \to b$ is an effective monomorphism if it has a cokernel pair and is the equalizer of its cokernel pair.

Note that an effective epimorphism (resp. monomorphism) is in particular a regular epimorphism (resp. monomorphism). In many categories of interest, the converse is also true.

Proposition: Let $f : a \to b$ be a morphism with a kernel pair $a \times_b a$. Then $f$ is a regular epimorphism if and only if it is an effective epimorphism.

Proof. One direction is clear. In the other, suppose $f$ is the coequalizer of some parallel pair of morphisms $g_1, g_2 : c \to a$. By assumption, $f$ has a kernel pair, so $g_1, g_2$ factor through it: that is, there is a map $h : c \to a \times_b a$ such that, letting $p_1, p_2 : a \times_b a \to a$ be the two projections, we have $p_i \circ h = g_i$.

It remains to show that $f$ is the coequalizer of $p_1, p_2$. But if $j : a \to d$ is another map such that $j \circ p_1 = j \circ p_2$, then

$\displaystyle j \circ g_1 = j \circ p_1 \circ h = j \circ p_2 \circ h = j \circ g_2$

so $j$ also coequalizes $g_1, g_2$, hence factors through $f$ by assumption. $\Box$

Dually, we obtain the following.

Proposition: Let $f : a \to b$ be a morphism with a cokernel pair $b \sqcup_a b$. Then $f$ is a regular monomorphism if and only if it is an effective monomorphism.

In particular, in a category with finite limits and colimits (which include all of the categories we discuss below), regular and effective coincide.

Example. The effective epimorphisms in $\text{Set}$ are precisely the surjective functions. This is just the statement that a surjective function $f : X \to Y$ is the quotient of $X$ by the equivalence relation on $X$ it induces. The effective monomorphisms are also precisely the injective functions. This follows from an explicit description of the cokernel pair of an injective function $f : X \to Y$, which is given by the disjoint union of $X$ together with two copies of $Y \setminus X$.

Example. The forgetful functor $\text{Top} \to \text{Set}$ preserves both limits and colimits, hence preserves kernel pairs, cokernel pairs, equalizers, and coequalizers. Equalizers are equipped with subspace topologies and coequalizers are equipped with quotient topologies. Consequently, the effective epimorphisms are precisely the quotient maps, and the effective monomorphisms are precisely the embeddings.

Example. The forgetful functor $\text{Grp} \to \text{Set}$ preserves limits but not colimits, hence preserves kernel pairs and equalizers but not cokernel pairs or coequalizers. However, it is still true that the effective epimorphisms are precisely the surjective homomorphisms; this is a form of the first isomorphism theorem. It also follows from the nontrivial fact we used about amalgamated free products earlier that the effective monomorphisms are precisely the injective homomorphisms.

Example. As above, the effective epimorphisms in $\text{Ring}$ are precisely the surjective homomorphisms, and as above, this is a form of the first isomorphism theorem (for rings). So we have successfully given a categorical definition of a surjective homomorphism of rings.

Effective monomorphisms do not behave as well. Recall that if $R$ is a commutative ring, then any nontrivial localization $f : R \to S^{-1} R$ is an epimorphism, hence has cokernel pair $S^{-1} R$. It follows that the equalizer of the cokernel pair of $f$ is the identity map $S^{-1} R \to S^{-1} R$, hence is not $f$, so $f$ is not an effective monomorphism although it is a monomorphism.

Isomorphisms

Previously we saw that an isomorphism is equivalently either a morphism which is both a monomorphism and a split epimorphism or a morphism which is both a split monomorphism and an epimorphism. This statement can be strengthened.

Theorem: The following conditions on a morphism $f : a \to b$ are equivalent:

1. $f$ is an isomorphism.
2. $f$ is a monomorphism and a regular epimorphism.
3. $f$ is a regular monomorphism and an epimorphism.

Proof. An isomorphism is both an equalizer and a coequalizer of any parallel pair of identical morphisms, so is in particular both a regular epimorphism and a regular monomorphism. Since being an isomorphism is a self-dual condition, it suffices to show that if $f$ is a monomorphism and a regular epimorphism then it is an isomorphism.

So suppose that $f$ is the coequalizer of a pair $g_1, g_2 : c \to a$ of morphisms. Then in particular $f \circ g_1 = f \circ g_2$, but by left cancellability $g_1 = g_2$. The coequalizer of any pair of identical morphisms is necessarily an isomorphism (since the identity map needs to factor through it and vice versa), so the conclusion follows. $\Box$

Corollary: Suppose $f : a \to b$ is an epimorphism but not an isomorphism. Then $f$ is not a regular monomorphism.

This generalizes the observation we made earlier about effective monomorphisms in $\text{Ring}$.

Strictly speaking, the above theorem does not imply the previous theorem until we know the following.

Proposition: A split epimorphism is regular.

Proof. Let $f : a \to b$ be a split epimorphism, hence there exists a section $g : b \to a$ (so $f \circ g = \text{id}_b$). The composite $g \circ f = m : a \to a$ is an idempotent. We claim that

$\displaystyle f = \text{coeq}(m, \text{id}_a)$.

To see this, let $h : a \to c$ be a morphism such that $h \circ m = h \circ \text{id}_a$. Then $h \circ g \circ f = h$, hence $h$ factors through $f$ as desired. $\Box$

Dually, we obtain the following.

Proposition: A split monomorphism is regular.

However, we obtain more than this: the proof above also shows that

$\displaystyle g = \text{eq}(m, \text{id}_a)$

hence both $f$ and $g$ can be recovered from the idempotent $m$ acting on $a$ alone. An idempotent $m$ such that the coequalizer and equalizer above both exist (equivalently, such that there exist $f : a \to b, g : b \to a$ such that $f \circ g = \text{id}_b, g \circ f = m$) is said to be split.

Note that $f$ and $g$ above are coequalizers resp. equalizers for purely diagrammatic reasons, so this remains true after an arbitrary functor has been applied; that is, the above are absolute coequalizers resp. equalizers.